Prove rational nums

Question

For all real number x : R(x) -> there exist two integers k, l such that x = k/l.

(i.e. x is a rational number)

Prove/Disprove: For all real number x : R(x) -> R(x+1)

My answer:

Let x be a real number.

Assume R(x).

Then R(x+1) = R(x) + R(1) # R(x) is rational and R(1) is rational because 1/1 = 1

Is that correct and complete? What is the more formal way of showing this proof?

Answer 1

If $x$ is rational, then $x = a/b$, where $a \in \mathbb{Z}$, $b \in \mathbb{Z}^*$.

$x+1 = \frac{a}{b} + 1 = \frac{a+b}{b}$. As $a+b \in \mathbb{Z}$, then $x+1$ is also rational.

Answer 2

Let $p/q$ be a rational, $q\neq0$. Then $p/q+1=(p+q)/q$ which is again a quotient of two integers, hence rational.

Answer 3

The set of all rationals with the usual addition and multiplication is a field; the number $1$ is rational; therefore, for every rational $x$ the number $x+1$ is again rational.

Answer 4

For all real number x : R(x) <-> there exist two integers k, l such that x = k/l.

Assuming $R(x)$ is true, then there exists integers $k, l$ such that $x=k/l$.   If so then $(x+1) = (k+l)/l$.   Now $(k+l)$ will be an integer, which means that there exists two integers $(k+l)$ and $l$ such that $(x+1)$ is their quotient.   That is $R(x+1)$.

$$\therefore R(x)\to R(x+1)$$

$\Box$